1
$\begingroup$

The integral is : $\displaystyle \lim_{n\rightarrow \infty} \int_{\mathbb{R}^2}e^{-(x^2+y^2)^n}dxdy$

How could I solve this integral?

I've tried polar coordinates, but then I get $\displaystyle \lim_{n\rightarrow \infty} \int_{0}^{\infty}e^{-r^{2n}}r2\pi \,dr$, and I don't know how to solve this...

Any help would be appreciated.

$\endgroup$
1
$\begingroup$

It's $$2\pi \lim_{n\to\infty }\int_0^\infty re^{-r^{2n}}\mathrm d r$$ and not what you wrote. If you make the substitution $u=r^{2n}$, you'll get $$2\pi \lim_{n\to\infty }\int_0^\infty \frac{1}{2n}u^{\frac{1}{n}-1}e^{-u}du=\pi\lim_{n\to\infty }\frac{1}{n}\Gamma\left(\frac{1}{n}\right)=\pi\lim_{n\to\infty }\Gamma\left(\frac{1}{n}+1\right)\underset{(*)}{=}\pi\Gamma(1)=\pi.$$

$(*) :$ By continuity of $\Gamma$.

$\endgroup$
5
  • $\begingroup$ by the way, how would I solve this using the dominated convergence theorem? $\endgroup$ – An old man in the sea. Nov 14 '15 at 17:46
  • $\begingroup$ I made a correction. You can't use directly converge dominated here (if you could, your integral would make $0$). $\endgroup$ – Surb Nov 14 '15 at 17:56
  • $\begingroup$ That's what I thought, but the exercise asked to use the CDT... I guess there is a mistake in the exercise. Thanks ;) $\endgroup$ – An old man in the sea. Nov 14 '15 at 17:59
  • $\begingroup$ Sorry for the inconvenience, but if you put [0,infinity[ instead of R, then shouldn't you adjust also the domain for theta and r accordingly? $\endgroup$ – An old man in the sea. Nov 14 '15 at 18:15
  • 1
    $\begingroup$ You are totally right, and I apologize for this mistakes. You have to take $\theta\in [0,\pi/2[$ (but $r$ is still positif). My correction was useless, then I let the first version (i.e. with $\theta\in [0,2\pi[$). $\endgroup$ – Surb Nov 14 '15 at 18:35
1
$\begingroup$

Substituting $R = r^2$, we want the limit as $n\to\infty$ of

$$\pi \int_0^\infty e^{-R^n} dR$$

Note that

$$\int_0^1 e^{-R^n} dR \leq \int_0^\infty e^{-R^n} dR \leq \int_0^1 1 \ dR + \int_1^\infty e^{-R^n} dR$$

Now squeeze.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.